outermeasure wrote:
kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).
Think about the line segment
. What must f be if you have
?
(The little circle thingy is dot product.)
Let x
≠ y.
Suppose f(y) = f(x) + grad f(x) o (y-x).
By strict convexity, for all 0<s<1,
f[(1-s)x+sy] < (1-s)f(x) + s f(y)
f[x+s(y-x)]< (1-s)f(x) + s [f(x) + grad f(x) o (y-x)]= f(x) + s [grad f(x) o (y-x)]
But where is the contradiction?